Solve-sin-2x-cos-2x-0-for-pi-lt-x-lt-pi-

Question Number 15975 by myintkhaing last updated on 16/Jun/17

S o l v e s i n 2 x + c o s 2 x = 0,

f o r - π < x < π .

Commented by tawa tawa last updated on 16/Jun/17

\sin 2 x + \cos 2 x = 0

Note : \sin 2 x = 2 sinxcosx and \cos 2 x = \cos^{2} x - \sin^{2} x

2 sinxcosx + \cos^{2} x - \sin^{2} x = 0

\cos^{2} x = 1 - \sin^{2} x

2 sinxcosx + 1 - \sin^{2} x - \sin^{2} x = 0

2 sinxcosx + 1 - {2 \sin}^{2} x = 0

∴ {2 \sin}^{2} x - 2 sinxcosx - 1 = 0

∴ {2 \sin}^{2} x - 2 sinx \sqrt{\cos^{2} x} - 1 = 0

∴ {2 \sin}^{2} x - 2 sinx \sqrt{1 - \sin^{2} x} - 1 = 0

sinx = p

∴ {2 p}^{2} - 2 p \sqrt{1 - p^{2}} - 1 = 0

∴ {2 p}^{2} - 1 = 2 p \sqrt{1 - p^{2}}

Square both sides

∴ {({2 p}^{2} - 1)}^{2} = {[2 p \sqrt{1 - p^{2}}]}^{2}

∴ ({2 p}^{2} - 1) ({2 p}^{2} - 1) = {4 p}^{2} (1 - p^{2})

∴ {4 p}^{4} - {2 p}^{2} - {2 p}^{2} + 1 = {4 p}^{2} - {4 p}^{4}

∴ {4 p}^{4} - {2 p}^{2} - {2 p}^{2} + 1 + {4 p}^{4} - {4 p}^{2} = 0

∴ {8 p}^{4} - {8 p}^{2} + 1 = 0

∴ 8 {(p^{2})}^{2} - {8 p}^{2} + 1 = 0

a = 8, b = - 8, c = 1

p^{2} = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}

p^{2} = \frac{- (- 8) \pm \sqrt{{(- 8)}^{2} - 4 (8) (1)}}{2 (8)}

p^{2} = \frac{8 \pm \sqrt{64 - 32}}{16}

p^{2} = \frac{8 \pm \sqrt{32}}{16}

p^{2} = 0.8536 or p^{2} = 0.1465

p = \pm \sqrt{0.8536} or p = \pm \sqrt{0.1465}

p = \pm 0.9239 or p = \pm 0.3828

sinx = \pm 0.9239 or sinx = \pm 0.3828

∴ x = \sin^{- 1} (0.9239) or x = \sin^{- 1} (- 0.9239) or x = \sin^{- 1} (0.3828) or x = \sin^{- 1} (- 0.3828)

∴ x = 67.5 or x = - 67.5 or x = 22.5 or x = - 22.5

To find other angle in the range given .

use : x = 180 n ° + {(- 1)}^{n} \sin^{- 1} (α), where α = \pm 67.5 or α = \pm 22.5

and n = 0, 1, 2 \dots Cancel anyone that is out of range .

Answered by mrW1 last updated on 16/Jun/17

\sqrt{2} (\frac{1}{\sqrt{2}} \sin 2 x + \frac{1}{\sqrt{2}} \cos 2 x) = 0

\sqrt{2} (\cos \frac{π}{4} \sin 2 x + \sin \frac{π}{4} \cos 2 x) = 0

\sqrt{2} \sin (2 x + \frac{π}{4}) = 0

\sin (2 x + \frac{π}{4}) = 0

\Rightarrow 2 x + \frac{π}{4} = k π, k \in Z

\Rightarrow x = \frac{1}{2} (k π - \frac{π}{4})

within (- π, π) :

x = - \frac{5 π}{8}, - \frac{π}{8}, \frac{3 π}{8}, \frac{7 π}{8}

Commented by myintkhaing last updated on 16/Jun/17

i n 2^{n d} l i n e, n o t s i n \frac{π}{2}

Commented by mrW1 last updated on 16/Jun/17

my typo, {it}^{'} s \frac{π}{4}

Commented by myintkhaing last updated on 16/Jun/17

s i n c o s + c o s s i n